Question

an aircraft is vertically above a point which is 10 km west and 15 km north from a control tower. If the aircraft is 4000 m above the ground, how far is it from the control tower.

Answer 1

Let the control tower be at the origin.
Assume north to be +y direction
              east to be +x direction
          upward to be +z direction
The aircraft is 10 km west(x = -10km) and 15 km north(y = +15km) and 4000m above the ground(z = +4000m = +4km).
So the coordinate of the aircraft is (-10,15,4).

$Distance\ from\ control\ tower\ or\ origin = \sqrt{(-10)^2+15^2+4^2}\ km \\distance= \sqrt{100+225+16}\ km \\distance = \sqrt{341}\ km \\ distance=\boxed{18.47\ km}$

Answer 2

Answer:

sqrt(341) = 18.47

Step-by-step explanation:

let c be the clocktower, a the plane and o as the origin. O is the point where NSWE intersect so 10km west from the origin will be AO , 15 km south of the origin which will be OC.

Now 4km below point A will be our point g, which is the gorund, below point C will be the exact location of the clocktower point T. draw an line from A to T whichll give us the hypotenuse. So AG is 4km and GT will be sqrt(15^2+10^2) = 18.03. which is the hypotenuse of AOC. so the distance from A to T, the distance of the plane to the clocktower will be sqrt(AG^2+GT^2) = 18.47